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borishaifa [10]
3 years ago
10

During an experiment, the percent yield of calcium chloride from a reaction was 85.22%. Theoretically, the expected amount shoul

d have been 113 grams. What was the actual yield from this reaction? CaCO3 + HCl → CaCl2 + CO2 + H2O
96.3 grams
99.0 grams
113 grams
121 grams
Chemistry
2 answers:
devlian [24]3 years ago
7 0

The formula to find yield is

(Actual Yield)/(Theorectical Yield) x100

Just do the math.

85.22% x 113 = 96.2986

Convert it to 3 significant figures

Ans: 96.3g

jonny [76]3 years ago
6 0

Answer : The actual yield of the reaction is, 96.3 g

Explanation :  Given,

Percent yield of the reaction = 82.22 %

Theoretical yield of the reaction = 113 g

The formula used for the percent yield will be :

\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

or,

\text{Actual yield}=\frac{(\text{Percent yield}\times \text{Theoretical yield})}{100}

Now put all the given values in this formula, we get:

\text{Actual yield}=\frac{(85.22\times 113)}{100}=96.3g

Therefore, the actual yield of the reaction is, 96.3 g

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Read 2 more answers
Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su
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Answer:

Volume = 746 L

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moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{175\ g}{101.543\ g/mol}

Moles_{copper(II)\ fluoride}= 1.7234\ mol

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Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

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Volume =\frac{1.7234}{0.00231}\ L

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3 years ago
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